Initial Stage of the Finite-Amplitude Cauchy–Poisson Problem

Abstract

The nonlinear Cauchy–Poisson problem for an incompressible inviscid fluid to start flowing under gravity is investigated analytically. The general nonlinear initial/boundary-value problem is formulated, including both an initial surface deflection and an initial velocity generated by a pressure impulse on the surface. Two subproblems are: (1) a finite-amplitude surface deflection released from rest; (2) the fluid is forced into motion by a pressure impulse on the initially horizontal surface. Solutions for these two subproblems are given to the leading order. One exact solution is given for the fully nonlinear initial-value problem, where a surface pressure impulse is applied on a surface with finite initial deflection. The concept of the highest non-breaking wave is illustrated by dipole acceleration fields at a state of gravitational release from rest. This is done for two phenomena: run-up of a non-breaking solitary wave on a sloping beach, and free nonlinear sloshing in an open container.

Appendix: A Family of Harmonic Functions

We show a recursive schemes for calculating higher orders of the harmonic functions $f_n(x,z)$ defined by their values at the boundary of the half-plane $z<0$

$$\begin{aligned} f_n(x,0) = \frac{1}{(1+x^2)^n}, \end{aligned}$$

where n is a positive integer. This family of multipole-type functions can be calculated recursively. First we introduce a source potential $\chi$ in the apex point (0, 1), which is a fictitious point located outside the fluid domain,

$$\begin{aligned} \chi = \frac{1}{2} \log (x^2 + (z-1)^2) \end{aligned}$$

and its gradient gives the first-order functions

$$\begin{aligned} f_1(x,z) = - \frac{\partial \chi }{\partial z} = \frac{1-z}{x^2 + (z-1)^2}, \end{aligned}$$

Each new order function is a linear combination of the z derivative of the previous function plus a linear combination of the lower order function.

$$\begin{aligned} 2 f_2= & {} f_1 + \frac{\partial f_1}{\partial z}, \\ 4 f_3= & {} \frac{1}{2} f_1 + 2 f_2 + \frac{\partial f_2}{\partial z}, \\ 6 f_4= & {} \frac{3}{8} f_1 + \frac{3}{4} f_2 + 3 f_3 +\frac{\partial f_3}{\partial z}, \\ 8 f_5= & {} \frac{5}{16} f_1 + \frac{1}{2} f_2 + f_3 + 4 f_4 + \frac{\partial f_4}{\partial z}, \\ 10 f_6= & {} \frac{35}{128} f_1 + \frac{25}{64} f_2 + \frac{5}{8} f_3 + \frac{5}{4} f_4 + 5 f_5 + \frac{\partial f_5}{\partial z}, \\ 12 f_7= & {} \frac{63}{256} f_1 + \frac{21}{64} f_2 + \frac{15}{32} f_3 + \frac{3}{4} f_4 + \frac{3}{2} f_5 + 6 f_6 +\frac{\partial f_6}{\partial z}, \\ 14 f_8= & {} \frac{231}{1024} f_1 + \frac{147}{512} f_2 + \frac{49}{128} f_3 + \frac{35}{64} f_4 + \frac{7}{8} f_5 + \frac{7}{4} f_6 + 7 f_7 + \frac{\partial f_7}{\partial z}, \\ 16 f_9= & {} \frac{429}{2048} f_1 + \frac{33}{128} f_2 + \frac{21}{64} f_3 + \frac{7}{16} f_4 + \frac{5}{8} f_5 + f_6 + 2 f_7 + 8 f_8 + \frac{\partial f_8}{\partial z}, \\ 18 f_{10}= & {} \frac{6435}{32768} f_1 + \frac{3861}{16384} f_2 + \frac{297}{1024} f_3 + \frac{189}{512} f_4 + \frac{63}{128} f_5 + \frac{45}{64} f_6 \\&+ \frac{9}{8} f_7 + \frac{9}{4} f_8 + 9 f_9 + \frac{\partial f_9}{\partial z}, \\ 20 f_{11}= & {} \frac{12155}{65536} f_1 + \frac{3575}{16384} f_2 + \frac{2145}{8192} f_3 + \frac{165}{512} f_4 + \frac{105}{256} f_5 + \frac{35}{64} f_6 + \frac{25}{32} f_7 \\&+ \frac{5}{4} f_8 + \frac{5}{2} f_9 + 10 f_{10}+ \frac{\partial f_{10}}{\partial z}, \end{aligned}$$

Note that these recursive formulas are valid in the entire half-plane $z \le 0$.